Please answer B, C, & D. The average amount of money spent for lunch per person in the collegecafeteria is $6.2 and the standard deviation is $2.54. Suppose that 43randomly selected lunch patrons are observed. Assume the distribution ofmoney spent is normal, and round all answers to 4 decimal places wherepossible.a. What is the distribution of X? X - N( 6.2b. What is the distribution of ¤? ¤ ~ N(c. For a single randomly selected lunch patron, find the probability thatthis patron's lunch cost is between $6.2091 and $6.6728.d. For the group of 43 patrons, find the probability that the averagelunch cost is between $6.2091 and $6.6728.e. For part d), is the assumption that the distribution is normalnecessary?YesO No o

Here, 43 samples are collected. b.) Hence, if the sample mean is denoted by x, then, x~N(μ,σ2n), where, μ=6.2, σ=2.54 and n=43. Thus, x~N(6.2,2.54243). c.) The probability that this patron's lunch cost is between $6.2091 and $6.6728 is X~N(6.2,2.542)⇒Z=X-6.22.54~N(0,1)P(6.2091<X<6.6728)=P(6.2091-6.22.54<X-6.22.54<6.6728-6.22.54)=P(0.0036<Z<0.186)=0.5738-0.5014=0.0724d.)The probability that the average patron's lunch cost is between $6.2091 and $6.6728 is x~N(6.2,2.54243)⇒Z1=43(x-6.2)2.54~N(0,1)P(6.2091<X<6.6728)=P(43(6.2091-6.2)2.54<43(x-6.2)2.54<43(6.6728-6.2)2.54)=P(0.0236<Z1<1.2197)=0.8887-0.5094=0.3793